A 9.00 kg hanging weight is connected by a string over a pulley to a 5.00 kg block that is sliding on a flat table.

Question

If the coefficient of kinetic friction is 0.230, find the tension in the string.

http://www.webassign.net/sb5/p5-49.gif

 

 

Answer 1

James did great job.

Here I'll show you a quick way.

9g - 5gμ = 14a , treating the two blocks as one system, then applying Newton's second law.

9g - T = 9a, applying Newton's second law for the 9 kg block.

Solve for T,

T = 9(g-a) = 9[g - (9g - 5gμ)/14] = 38.745 N, where g = 9.8 m/s^2, and μ = 0.230.

Answer 2

If we start with free-body diagrams for each of the two masses(which I unfortunately cannot reproduce here) we find that the forces acting on m₁ (the 9 kg weight) are tension T in the positive y direction and weight (m₁g) in the negative y direction. Newton's second law F=ma for this block then reads:

T-m₁g=m₁a₁                             <---equation 1

where a₁ is the acceleration of m₁

The forces acting on m₂ (the 5 kg block) are tension T in the negative x direction, friction f in the positive x direction, a normal force N in the positive y direction, and weight in the negative y direction. I chose to orient my diagram so that the movement of each mass will be in a negative direction, assuming we agree to use the traditional x and y coordinate system. Using the forces in the y directions and F=ma yields N-m₂g=0 (since the block is not accelerating vertically) or N=m₂g. This will be useful for the frictional force. The forces in the x direction combine with Newton's second to give:

-T+f=m₂a₂                             <---equation 2

using the definition of the frictional force f=uN and the fact that a₁=a₂ (since we assume the rope does not stretch or break) we can rewrite equation 2:

-T+uN=m₂a₁                         <---equation 3

Using equations 1 and 3 we have 2 equations with two unknowns(T and a₁). For me the simplest way to solve for the tension is to divide one equation by the other and thereby eliminate a₁ altogether. What is left is one equation with one unknown variable. Let me illustrate:

   T-m₁g  = m₁a₁                      <---equation 1

-----------   -------

 -T+um₂g = m₂a₁                      <---equation 3

It looks a bit awkward here but the point is that we now have one equation consisting of fractions on either side. On the right side the a₁'s cancel leaving us with the tension as our only unknown. If you then cross multiply, group all the terms with T on side and all the other terms on the other side, factor out and solve for T you get:

T=(um₂gm₁ + m₁m₂g)/(m₁ + m₂)

(apologies for skipping the algebra)

using the numbers you gave in the question I get a result of 38.8[Newtons]